Let $f(x)=x^6-3x^5$. On which intervals is $f$ decreasing? Choose 1 answer: Choose 1 answer: (Choice A) A $x<\dfrac52$ only (Choice B) B $0<x<\dfrac52$ only (Choice C) C $x>0$ only (Choice D) D $x<0$ only (Choice E) E The entire domain of $f$
We can analyze the intervals where $f$ is increasing/decreasing by looking for the intervals where its derivative $f'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $f$ is $f'(x)=3x^4(2x-5)$. $f'(x)=0$ for $x=0,\dfrac52$. Since $f'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$ and $x=\dfrac52$. $f$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $ x<0$ $ 0<x<\frac{5}{2}$ $\frac{5}{2}$ $ x>\frac{5}{2}$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $x<0$ $x=-1$ $f'(-1)=-21<0$ $f$ is decreasing $\searrow$ $0<x<\dfrac{5}{2}$ $x=1$ $f'\left(1\right)=-9<0$ $f$ is decreasing $\searrow$ $x>\dfrac{5}{2}$ $x=3$ $f'(3)=243>0$ $f$ is increasing $\nearrow$ In conclusion, $f$ is decreasing over the interval $x<\dfrac52$.